# Math Help - Application problems are really starting to annoy me...

1. ## Application problems are really starting to annoy me...

For a certain model of car the distance d required to stop the vehicle if it is traveling at v mi/h is given by the formula

$d = v + \frac{v^2}{20}$

where d is measured in feet. Kerry wants her stopping distance not to exceed 240 ft. At what range of speeds can she travel?
I have made a couple attempts at setting up this problem, but both have failed miserably so far. So some help setting it up will be appreciated.

Also, since I am having so much trouble with application problems, is there any tips you guys can give me on how I should attack future math problems such as these? Thanks.

2. Hello,
Originally Posted by mathgeek777
I have made a couple attempts at setting up this problem, but both have failed miserably so far. So some help setting it up will be appreciated.

Also, since I am having so much trouble with application problems, is there any tips you guys can give me on how I should attack future math problems such as these? Thanks.
The main tip I can give you for this is that you always must translate the sentences given in the text.

Here : "she wants her stopping distance" << this is d "not to exceed 240ft", that is d<240.

So you have to find v such that d = v + v²/20 < 240.

Advice : be careful, speed is in mi/h and distance in ft in the formula. So don't bother changing the units to be the same or wondering about them your result will automatically be in mi/h.

Edit : woah... 28th post

3. Originally Posted by mathgeek777
I have made a couple attempts at setting up this problem, but both have failed miserably so far. So some help setting it up will be appreciated.

Also, since I am having so much trouble with application problems, is there any tips you guys can give me on how I should attack future math problems such as these? Thanks.
Let $d=240$

The equation then becomes $240=v+\frac{v^2}{20}\implies v^2+20v-4800=0$

Applying the quadratic formula, we see that $v=\frac{-20\pm\sqrt{400+19200}}{2}\implies v=\frac{-20\pm140}{2}$

This implies that $v=-80~or~v=60$

We take the positive value.

So the maximum speed she can travel in order to break a distance of 240 ft would be 60 mph.

I hope this makes sense!

--Chris

EDIT: Shucks...Moo beats me again...I guess the early cow gets the grass

4. Thanks for the help guys. Maybe I should carefully look at a problem and figure out what I have and don't have first, rather than jump straight into it right away only to run into trouble.